Display data from MySql

[slj90]

I'm trying to display data from a MySQL table using PHP.

 

At the minute it is repeating the frist row 23 times, there are 23 rows in the table. What do I need to add to my code to make it display each row?

 

$result=mysqli_query($con,"SELECT item_item_title, item_username FROM items");
$row = mysqli_fetch_array($result);
$item_item_title = $row[item_item_title];
$item_username = $row[item_username];

if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        echo $item_item_title . "     -     " . $item_username . "<br>";
    }
} else {
    echo "0 results";
}

Thanks

[Jacques1]

You need to fix the logic of your code. Right now, you fetch the first row (which may not even exist) and extract the item title and username. Then you iterate over the remaining rows, but you just keep echoing the title and username of the first row.

 

Get rid of the extra code for the first row and extract the title and name for each row within the loop.

[mlukac89]

Try change to this

$result=mysqli_query($con,"SELECT item_item_title, item_username FROM items");

if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        echo $row[item_item_title] . "     -     " . $row[item_username] . "<br>";
    }
} else {
    echo "0 results";
}

Edited October 12, 2015 by mlukac89

[Ani123]

Hey,try this code:

$query="SELECT * FROM MY_TABLE";
$results = mysql_query($query);

while ($row = mysql_fetch_array($results)) {
   echo '<tr>';
   foreach($row as $field) {
       echo '<td>' . htmlspecialchars($field) . '</td>';
   }
   echo '</tr>';
}

[benanamen]

@Ani123, No, no, no! You dont loop through the data twice.